Physics, asked by amark3468, 1 year ago

Moment of inertia of triangular plate about its centroid

Answers

Answered by taranpreetkaur3
1

)

eight of triangle is h = ¼ L√3

if triangle is one of its edge on below position, so the height of center is ⅔ h.

I = ∫ r² dA

x = yL/h =

dA = x dy = (yL/h) dy

moment inertia with respect to X-axis is

Ix = ∫ y² dA = ∫ y² (yL/h) dy = (L/h) ∫ y³ dy

Ix = (L/h) (¼) y⁴ which y from 0 to h

Ix = (L/h) (h⁴ - 0) = ¼ Lh³

according to parallel axis theorem, calculate moment inertia w.r.t X-axis through the center of mass is,

Ix = Ixo + Ay²

¼ Lh³ = Ixo + (½ Lh) (⅔h)²

Ixo = (1/36) Lh³

Let's we calculate moment inertia with respect to Y-axis through the center of mass is,

Iy = ∫ x² dA

dA = y dx

y = -(x - h)

Iyo = ∫ x² dA = 2 ∫ x² (-(x - h)) dx from x = 0 to x = ½ L

Iyo = 2 ∫ (-x³ + x²h) dx = -½ x⁴ + ⅔ x³h from x = 0 to x = ½ L

Iyo = (-1/32) L⁴ + (1/12) L³h

notify that Y-axis through the center of mass.

according to Pythagoras theorem,

r² = x² + y²

∫ r² dA = ∫ x² dA + ∫ y² dA

Ir = Iyo + Ixo

Ir = (-1/32) L⁴ + (1/12) L³h + (1/36) Lh³

but h = ¼ L√3

Ir = L⁴(17√3 - 24)/768

moment of inertia of an equilateral triangular plate of length L about its centre is Ir = L⁴(17√3 - 24)/768

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Answered by akanksha2614
0

Answer:

Let the mass density be ρkg/m

2

Let the side PQ=RS=a & QR=PS=b (From fig a>b)

Mass of rectangle is m= ρab

Therefore Moment of Inertia of Rectangle about its center = m

12

a

2

+b

2

Distance of P point from center of rectangle is

2

a

2

+b

2

Therefore Moment of Inertia of Rectangle about P, I= m

12

a

2

+b

2

+m

4

a

2

+b

2

=m

3

a

2

+b

2

Mass of triangle PQR=

2

m

=

2

ρab

Moment of Inertia of Triangle PQR about its centroid = ρ

12

ab

3

+ba

3

=m

12

a

2

+b

2

Distance of point P from centroid =

(

3

2a

)

2

+(

3

b

)

2

Moment of Inertia of Triangle PQR about P= m

12

a

2

+b

2

+

2

m

{(

3

2a

)

2

+(

3

b

)

2

}=

2

m

18

11a

2

+5b

2

>

2

m

18

6a

2

+6b

2

>

2

I

(As a>b)

Distance of point R from centroid =

(

3

2b

)

2

+(

3

a

)

2

Moment of Inertia of TrianglePQR about P= m

12

a

2

+b

2

+

2

m

{(

3

a

)

2

+(

3

2b

)

2

}=

2

m

18

5a

2

+11b

2

As a>b therefore

2

m

18

5a

2

+11b

2

can be less than or equal to or greater than

2

I

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